The relation “is parallel to” on a set S of all straight lines in a plane is : -Maths 9th

The relation “is parallel to” on a set S of all straight lines in a plane is : -Maths 9th
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(d) An equivalence relationLet R = {(x, y) : line x is parallel to line y, x y ∈ set of coplanar straight lines}. • Every line is parallel to itself. So, if x ∈S, then (x, x) ∈R ⇒ R is reflexive • If (x, y) ∈R ⇒ x | | y ⇒ y | | x ⇒ (y, x) ∈R ⇒ R is symmetric • (x, y) ∈R and (y, z) ∈R ⇒ x | | y and y | | z ⇒ x | | z ⇒ (x, z) ∈R ⇒ R is transitive ∴ R being reflexive, symmetric and transitive, it is an equivalence relation.
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Related questions

A straight line is parallel to the lines 3x – y – 3 = 0 and 3x – y + 5 = 0 and lies between them. -Maths 9th

Description : A straight line is parallel to the lines 3x – y – 3 = 0 and 3x – y + 5 = 0 and lies between them. -Maths 9th

Last Answer : (c) 3x - 4y + 15 = 0 Let (m > 0) be the gradient (slope) of the required line. Then, Equation of any line through (-5, 0) having slope = m is y - 0 = m(x - (-5)) or mx - y + 5m = 0 ...(i) Its ... (rac{3}{4}\) (∵ m is +ve)∴ Required equation: y = \(rac{3}{4}\) (x + 5)⇒ 3x - 4y + 15 = 0.

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Description : Show that the relation ‘≅’ congruence on the set of all triangles in Euclidean plane geometry is an equivalence relation. -Maths 9th

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If m parallel lines in a plane are intersected by a family of n parallel lines, find the number of parallelograms formed? -Maths 9th

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PQ and RS are two equal and parallel line segments.Any points M not lying on PQ or RS is joined to Q and S and lines through P parallel to SM meet at N.Prove that line segments MN and PQ are equal and parallel to each other. -Maths 9th

Description : PQ and RS are two equal and parallel line segments.Any points M not lying on PQ or RS is joined to Q and S and lines through P parallel to SM meet at N.Prove that line segments MN and PQ are equal and parallel to each other. -Maths 9th

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In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Description : In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Last Answer : Given X and Y are the mid-points of AC and AB respectively. Also, QP|| BC and CYQ, BXP are straight lines. To prove ar (ΔABP) = ar (ΔACQ) Proof Since, X and Y are the mid-points of AC and AB respectively. So, ... ar (ΔBYX) + ar (XYAP) = ar (ΔCXY) + ar (YXAQ) ⇒ ar (ΔABP) = ar (ΔACQ) Hence proved.

In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Description : In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

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Draw the graph of the equation represented by a straight Line which is parallel to the X-axis and at a distance 3 units below it. -Maths 9th

Description : Draw the graph of the equation represented by a straight Line which is parallel to the X-axis and at a distance 3 units below it. -Maths 9th

Last Answer : Any straight line parallel to X-axis in negative direction of Y-axis is given by y = - k, where k is the distance of the line from the X-axis. Here, k = 3. Therefore, the equation of the line is y = -3. To ... , plot the points (1,-3), (2, -3) and (3, -3) and join them. This is the required graph.

Draw the graph of the equation represented by a straight Line which is parallel to the X-axis and at a distance 3 units below it. -Maths 9th

Description : Draw the graph of the equation represented by a straight Line which is parallel to the X-axis and at a distance 3 units below it. -Maths 9th

Last Answer : Any straight line parallel to X-axis in negative direction of Y-axis is given by y = - k, where k is the distance of the line from the X-axis. Here, k = 3. Therefore, the equation of the line is y = -3. To ... , plot the points (1,-3), (2, -3) and (3, -3) and join them. This is the required graph.

Draw the graph of the equation represented by the straight line which is parallel to the x-axis and 3 units above it. -Maths 9th

Description : Draw the graph of the equation represented by the straight line which is parallel to the x-axis and 3 units above it. -Maths 9th

Last Answer : Solution :- Any straight line parallel to x-axis is given by y = a, where a is the distance of the line from the x-axis. Here a = 3. Therefore the equation of the line is y = 3. To draw the graph of this equation plot the points (0, 3) and (3, 3) and join them.

What is the name of the horizontal and the vertical lines drawn to determine the position of any point in the Cartesian plane? -Maths 9th

Description : What is the name of the horizontal and the vertical lines drawn to determine the position of any point in the Cartesian plane? -Maths 9th

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Let N be the set of natural numbers. Describe the following relation in words giving its domain -Maths 9th

Description : Let N be the set of natural numbers. Describe the following relation in words giving its domain -Maths 9th

Last Answer : The given relation stated in words is R = {(x, y) : x is the fourth power of y; x ∈ N, y ∈ {1, 2, 3, 4}}.

Choose the correct option. The relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} on a set A = {1, 2, 3} is -Maths 9th

Description : Choose the correct option. The relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} on a set A = {1, 2, 3} is -Maths 9th

Last Answer : R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} on set A = {1, 2, 3}. (i) Since (1, 1), (2, 2), (3, 3) ∈ R ⇒ R is reflexive. (ii) (1, 2) ∈ R but (2, 1) ∉ R ⇒ R is not symmetric. (iii) (1, 2) ∈ R, (2, 3) ∈ R and (1, 3) ∈R ⇒ R is transitive. ∴ Option (a) is the right answer.

Show that the relation R in the set A of all the books in a library of a school given by R = {(x, y): x and -Maths 9th

Description : Show that the relation R in the set A of all the books in a library of a school given by R = {(x, y): x and -Maths 9th

Last Answer : Given: A = {All books in a library of a school} R = {(x, y): x and y have the same number of pages} Reflexivity: (x, x) ∈ R ⇒ R is reflexive on A Symmetric: Since books x and y have the same number of pages ... and (y, z) ∈ R ⇒ (x, z) ∈ R ⇒ R is transitive on A. Hence, R is an equivalence relation.

Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b): b = a + 1} is reflexive, symmetric or transitive. -Maths 9th

Description : Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b): b = a + 1} is reflexive, symmetric or transitive. -Maths 9th

Last Answer : Reflexive: R = {(a, b) : b = a +1} = {(a, a + l) : a, a + 1∈{l, 2, 3, 4, 5, 6}} = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)} ⇒ R is not reflexive since (a, a) ∉R for all a. Symmetric: R is not symmetric as (a ... as (a, b) ∈ R and (b, c) ∈ R but (a, c) ∉ R e.g., (1, 2) ∈ R (2, 3) ∈ R but (1, 3) ∉R

The relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is: -Maths 9th

Description : The relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is: -Maths 9th

Last Answer : (c) Symmetric onlyAs (1, 1), (2, 2) ∉ R so R is not reflexive. A relation a R b is said to symmetric if (a, b) ∈ R ⇒ (b, a) ∈ R. Here (1, 2) ∈ R and (2, 1) ∈ R so it is symmetric. Also, as (1, 2) ∈ R, but (2, 3), (1, 3) ∉ R, so R is not transitive.

Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 12), (3, 6)} be a relation on set A = {3, 6, 9, 12}. The relation is -Maths 9th

Description : Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 12), (3, 6)} be a relation on set A = {3, 6, 9, 12}. The relation is -Maths 9th

Last Answer : (c) Reflexive and transitive onlyHere A = {3, 6, 9, 12} and R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 12), (3, 6)} ∵ (3, 3), (6, 6), (9, 9), (12, 12) all belong to R ⇒ {(a, a): ∈R V a ∈A} ... 12) ∈ R and (3,12) ∈ R ⇒ (3, 12) ∈ R (x, y) ∈ R, (y, z) ∈ R ⇒ (x, z) ∈ R Hence R is transitive.

Which of the following is an equivalence relation defined on set A = {1, 2, 3} -Maths 9th

Description : Which of the following is an equivalence relation defined on set A = {1, 2, 3} -Maths 9th

Last Answer : (c) {(1, 1), (2, 2), (3, 1), (1, 3), (3, 3)} Option (c) satisfies all the conditions of an equivalence relation. (1, 1), (2, 2), (3, 3) ∈ R ⇒ (a, a) ∈ R V a ∈ A ⇒ R is reflexive (3, 1) ∈ R and (1, 3) ∈ R ... R and (1, 1) ∈ R ⇒ (a, b) ∈ R, (b, c) ∈ R ⇒ (a, c) ∈ R V a, b, c ∈ A ⇒ R is transitive.

Given the relation R = {(1, 2), (2, 3)} on the set A = {1, 2, 3}, -Maths 9th

Description : Given the relation R = {(1, 2), (2, 3)} on the set A = {1, 2, 3}, -Maths 9th

Last Answer : (d) 7A = {1, 2, 3}. R = {(1, 2), (2, 3)} To make R an equivalence relation, it should be: (i) Reflexive: So three more ordered pairs (1, 1), (2, 2), (3, 3) should be added to R to make it ... 2, 1), (3, 2), (1, 3), (3, 1)} is an equivalence relation. So minimum 7 ordered pairs are to be added.

Let R be a relation defined on the set A of all triangles such that R = {(T1, T2) : T1 is similar to T2}. Then R is -Maths 9th

Description : Let R be a relation defined on the set A of all triangles such that R = {(T1, T2) : T1 is similar to T2}. Then R is -Maths 9th

Last Answer : (d) An equivalence relation.Every triangle is similar to itself, so (T1, T1) ∈ R ⇒ R is reflexive. (T1, T2) ∈ R ⇒ T1 ~ T2 ⇒T2 ~ T1, ⇒ (T2, T1) ∈ R ⇒ R is symmetrictransitive. ∴ R is an equivalence relation.

On a set N of all natural numbers is defined the relation R by a R b iff the GCD of a and b is 2, then R is -Maths 9th

Description : On a set N of all natural numbers is defined the relation R by a R b iff the GCD of a and b is 2, then R is -Maths 9th

Last Answer : (c) Symmetric only Let a ∈N. Then (a, a) ∉R as the GCD of a' and a' is a' not 2. R is not reflexive Let a, b ∈N. Then, (a, b) ∉R ⇒ GCD of a' and b' is 2 ⇒ GCD of b' and a' is 2 ⇒ (b, a) ∈R ∴ R ... , let a = 4, b = 10, c = 12 GCD of (4, 10) = 2 GCD of (10, 12) = 2 But GCD of (4, 12) = 4.

On the set R of all real numbers, a relation R is defined by R = {(a, b) : 1 + ab > 0}. Then R is -Maths 9th

Description : On the set R of all real numbers, a relation R is defined by R = {(a, b) : 1 + ab > 0}. Then R is -Maths 9th

Last Answer : (a) Reflexive and symmetric only(a, a) ∈ R ⇒ 1 + a . a = 1 + a2 > 0 V real numbers a ⇒ R is reflexive (a, b) ∈ R ⇒ 1 + ab > 0 ⇒ 1 + ba > 0 ⇒ (b, a) ∈ R ⇒ R is symmetricWe observe that \(\big(1,rac{1}{2}\big) ... }{2},-1\big)\) ∈ Rbut (1, - 1) ∉ R as 1 + 1 (-1) = 0 \( ot>\) 0 ⇒ R is not transitive.

If R is a relation defined on the set of natural numbers N such that (a, b) R (c, d) if and only if a + d = b + c, then R is -Maths 9th

Description : If R is a relation defined on the set of natural numbers N such that (a, b) R (c, d) if and only if a + d = b + c, then R is -Maths 9th

Last Answer : (d) An equivalence relationWe can check the given properties as follows: Reflexive: Let (a, b) ∈ N x N. Then (a, b) ∈ N ⇒ a + b = b + a (Communtative law of Addition) ⇒ (a, b) R (b, a) ⇒ (a, b) R (a, ... , f) ⇒ (a, b) R (e, f) on N x N so R is transitive.Hence R is an equivalence relation on N N.

Let R be a relation on the set of integers given by a = 2^k .b for some integer k. Then R is -Maths 9th

Description : Let R be a relation on the set of integers given by a = 2^k .b for some integer k. Then R is -Maths 9th

Last Answer : (c) equivalence relationGiven, a R b = a = 2k .b for some integer. Reflexive: a R a ⇒ a = 20.a for k = 0 (an integer). True Symmetric: a R b ⇒ a = 2k b ⇒ b = 2-k . a ⇒ b R a as k, -k are both ... = 2k1 + k2 c, k1 + k2 is an integer. ∴ a R b, b R c ⇒ a R c True ∴ R is an equivalence relation.

Let R be a relation on the set N, defined by {(x, y) : 2x – y = 10} then R is -Maths 9th

Description : Let R be a relation on the set N, defined by {(x, y) : 2x – y = 10} then R is -Maths 9th

Last Answer : (a) ReflexiveGiven, {(\(x\), y) : 2\(x\) – y = 10} Reflexive, \(x\) R \(x\) = 2\(x\) – \(x\) = 10 ⇒ \(x\) = 10 ⇒ y = 10 ∴ Point (10, 10) ∈ N ⇒ R is reflexive.

Let A = {1, 2, 3, 4}, B = {5, 6, 7, 8}. Then R = {(1, 5), (1, 7), (2, 6)} is a relation from set A to B defined as : -Maths 9th

Description : Let A = {1, 2, 3, 4}, B = {5, 6, 7, 8}. Then R = {(1, 5), (1, 7), (2, 6)} is a relation from set A to B defined as : -Maths 9th

Last Answer : (d) R = {(a, b) : b/a is odd} Since (2, 6) ∈R, the relation a and b are odd does not exist. Since (1, 5) and (1, 7) ∈R, the relation a and b are even does not exist. None of the ... \(rac{7}{1}\) = 7, \(rac{6}{2}\) = 3, quotients being all odd numbers, the relation b/a is odd exists.

Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Then R is -Maths 9th

Description : Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Then R is -Maths 9th

Last Answer : (b) Reflexive and transitive but not symmetricLet A = {1, 2, 3, 4} • ∵ (1, 1), (2, 2), (3, 3) and (4, 4) ∈R ⇒ R is reflexive • ∵ (1, 2) ∈ R but (2, 1) ∉ R ; (1, 3) ∈ R and (3, 1) ∉ R ; (3, 2) ∈R and (2, 3) ∉ R ⇒ R is not symmetric • (1, 3) ∈ R and (3, 2) ∈ R and (1, 2) ∈ R ⇒ R is transitive.

If R is a relation on a finite set A having n elements, then the number of relations on A is -Maths 9th

Description : If R is a relation on a finite set A having n elements, then the number of relations on A is -Maths 9th

Last Answer : (d) \(2^{n^2}\)Set A has n elements ⇒ n(A) = n ⇒ A × A has n × n = n2 elements ∴ Number of relations on A = Number of subsets of A × A = \(2^{n^2}\)

The relation ‘is less than’ on a set of natural numbers is -Maths 9th

Description : The relation ‘is less than’ on a set of natural numbers is -Maths 9th

Last Answer : (c) Only transitiveLet N be the set of natural numbers. Then R = {(a, b) : a < b, a, b ∈N}A natural number is not less than itself ⇒ (a, a)∉R where a ∈N ⇒ R is not reflexive V a, b ∈N, (a, b) ∈R ⇒ a < b \( ot\ ... ∈N, (a, b) ∈R and (b, c) ∈R ⇒ a < b and b < c ⇒ a < c (a, c) ∈R ⇒ R is transitive.

EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Description : EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Last Answer : AB || CD and a transversal EF intersects them ∴ ∠EGB = ∠GHD ( Corresponding Angles) ⇒ 2 ∠EGM = 2 ∠GHL ∵ GM and HL are the bisectors of ∠EGB and ∠EHD respectively. ⇒ ∠EGM = ∠GHL But these angles form a pair of equal corresponding angles for lines GM and HL and transversal EF. ∴ GM || HL.

l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Description : l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Last Answer : Though E, draw a line parallel to p intersecting L at G and n at H respectively. Since l | | m ⇒ AG | | BE and AB | | GE [by construction] ∴ Opposite sides of quadrilateral AGEB are ... ∠DGE = ∠FHE [alternate interior angles] By ASA congruence axiom, we have △DEG ≅ △FEH Hence, DE = EF

AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Description : AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Last Answer : Given In the figure l || m, AP and BQ are the bisectors of ∠EAB and ∠ABH, respectively. To prove AP|| BQ Proof Since, l || m and t is transversal. Therefore, ∠EAB = ∠ABH [alternate interior ... ∠PAB and ∠ABQ are alternate interior angles with two lines AP and BQ and transversal AB. Hence, AP || BQ.

Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other. -Maths 9th

Description : Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other. -Maths 9th

Last Answer : Given Two lines m and n are parallel and another two lines p and q are respectively perpendicular to m and n. i.e., p ⊥ m, p ⊥ n, q ⊥ m, q ⊥ n To prove p||g Proof Since, ... of interior angles on the same side of the transversal is supplementary, then the two lines are parallel. Hence, p||g.

A transversal intersects two parallel lines. -Maths 9th

Description : A transversal intersects two parallel lines. -Maths 9th

Last Answer : Given Two lines AB and CD are parallel and intersected by transversal t at P and 0, respectively. Also, EP and FQ are the bisectors of angles ∠APG and ∠CQP, respectively.

If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form -Maths 9th

Description : If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form -Maths 9th

Last Answer : According to question the bisectors of the angles APQ, BPQ, CQP and PQD form

EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Description : EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Last Answer : AB || CD and a transversal EF intersects them ∴ ∠EGB = ∠GHD ( Corresponding Angles) ⇒ 2 ∠EGM = 2 ∠GHL ∵ GM and HL are the bisectors of ∠EGB and ∠EHD respectively. ⇒ ∠EGM = ∠GHL But these angles form a pair of equal corresponding angles for lines GM and HL and transversal EF. ∴ GM || HL.

l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Description : l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Last Answer : Though E, draw a line parallel to p intersecting L at G and n at H respectively. Since l | | m ⇒ AG | | BE and AB | | GE [by construction] ∴ Opposite sides of quadrilateral AGEB are ... ∠DGE = ∠FHE [alternate interior angles] By ASA congruence axiom, we have △DEG ≅ △FEH Hence, DE = EF

AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Description : AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Last Answer : Given In the figure l || m, AP and BQ are the bisectors of ∠EAB and ∠ABH, respectively. To prove AP|| BQ Proof Since, l || m and t is transversal. Therefore, ∠EAB = ∠ABH [alternate interior ... ∠PAB and ∠ABQ are alternate interior angles with two lines AP and BQ and transversal AB. Hence, AP || BQ.

Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other. -Maths 9th

Description : Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other. -Maths 9th

Last Answer : Given Two lines m and n are parallel and another two lines p and q are respectively perpendicular to m and n. i.e., p ⊥ m, p ⊥ n, q ⊥ m, q ⊥ n To prove p||g Proof Since, ... of interior angles on the same side of the transversal is supplementary, then the two lines are parallel. Hence, p||g.

A transversal intersects two parallel lines. -Maths 9th

Description : A transversal intersects two parallel lines. -Maths 9th

Last Answer : Given Two lines AB and CD are parallel and intersected by transversal t at P and 0, respectively. Also, EP and FQ are the bisectors of angles ∠APG and ∠CQP, respectively.

If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form -Maths 9th

Description : If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form -Maths 9th

Last Answer : According to question the bisectors of the angles APQ, BPQ, CQP and PQD form

Does Euclid's fifth postulate imply the existence of parallel lines?Explain. -Maths 9th

Description : Does Euclid's fifth postulate imply the existence of parallel lines?Explain. -Maths 9th

Last Answer : Solution :-

A transversal intersects two lines in such a way that the two interior angle on the same side of transversal are equal.Will the two lines always be parallel? -Maths 9th

Description : A transversal intersects two lines in such a way that the two interior angle on the same side of transversal are equal.Will the two lines always be parallel? -Maths 9th

Last Answer : Solution :- The two lines will not be always parallel as the sum of the two equal angles will not always be 180°. Lines will be parallel when each of the equal angles is equal to 90°.

In Fig.6.5,which of the two lines are parallel? -Maths 9th

Description : In Fig.6.5,which of the two lines are parallel? -Maths 9th

Last Answer : Solution :- l||m, because angles on the same side of the transversal are supplementary, i.e., 128° +52° = 180°. Therefore p is not parallel to q, because 105° + 74° = 179°.

In Fig.6.6, find the value of x for which the lines l and m are parallel. -Maths 9th

Description : In Fig.6.6, find the value of x for which the lines l and m are parallel. -Maths 9th

Last Answer : Solution :-

For what value of x will the lines l and m be parallel to each other ? -Maths 9th

Description : For what value of x will the lines l and m be parallel to each other ? -Maths 9th

Last Answer : Solution :-

If a transversal intersects two parallel lines, prove that the bisectors of any pair of corresponding angles so formed are parallel. -Maths 9th

Description : If a transversal intersects two parallel lines, prove that the bisectors of any pair of corresponding angles so formed are parallel. -Maths 9th

Last Answer : Solution :-

Two parallel lines l and m are intersected by a transversal p (see Fig. 8.46). Show that the quadrilateral formed by the bisectors of interior angles is a rectangle. -Maths 9th

Description : Two parallel lines l and m are intersected by a transversal p (see Fig. 8.46). Show that the quadrilateral formed by the bisectors of interior angles is a rectangle. -Maths 9th

Last Answer : Solution :-

l,m and n are three parallel lines intersected by transversal p and q such that l,m and n cut-off equal intersepts AB and BC on p (Fig.8.55). Show that l,m and n cut - off equal intercepts DE and EF on q also. -Maths 9th

Description : l,m and n are three parallel lines intersected by transversal p and q such that l,m and n cut-off equal intersepts AB and BC on p (Fig.8.55). Show that l,m and n cut - off equal intercepts DE and EF on q also. -Maths 9th

Last Answer : Given:l∥m∥n l,m and n cut off equal intercepts AB and BC on p So,AB=BC To prove:l,m and n cut off equal intercepts DE and EF on q i.e.,DE=EF Proof:In △ACF, B is the mid-point of ... a triangle, parallel to another side, bisects the third side. Since E is the mid-point of DF DE=EF Hence proved.

Whether the pair of given lines are parallel or not give reason. -Maths 9th

Description : Whether the pair of given lines are parallel or not give reason. -Maths 9th

Last Answer : Where are the lines please tell first